CBSE Class 12 Chemistry Coordination Compounds NEET JEE Questions Set 03

Single Correct Answer

Question 1. Hybridization, shape and magnetic moment of K₃[Co(CO₃)₃] is (Odisha JEE 2008)
(a) d²sp³, octahedral, 4.9 B.M.
(b) sp³d², octahedral, 4.9 B.M.
(c) dsp², square planar, 4.9 B.M.
(d) sp³, tetrahedral, 4.9 B.M.
Answer:
Ans. (b)
In each case, magnetic moment is 4.9 B.M. which shows that each of them has 4 unpaired electrons. Further, in the given complex, oxidation state of cobalt is +3. Hence, it will have 3d⁶ configuration, i.e., four unpaired electrons remain in 3d and no 3d orbital is free.
Thus, hybridisation possible is sp³d² (4s, 4p, 4d), i.e., outer orbital (high spin) octahedral complex.

Teacher's Note:
1. Each CO₃²⁻ is didentate, so three of them give coordination number 6 (octahedral), not 3.

Question 2. A magnetic moment of 1.73 BM will be shown by one among the following : (AIPMT 2013)
(a) [Cu(NH₃)₄]²⁺
(b) [Ni(CN)₄]²⁻
(c) TiCl₄
(d) [CoCl₆]⁴⁻
Answer:
Ans. (a)
In [Cu(NH₃)₄]²⁺, Cu²⁺ = 3d⁹, i.e., it contains one unpaired electron. Magnetic moment,
\( \mu = \sqrt{n(n+2)} \) BM
∴ \( 1.73 = \sqrt{n(n+2)} \) or n = 1
So the complex containing one unpaired electron is [Cu(NH₃)₄]²⁺.

Teacher's Note:
1. Remember the common values: 1.73 BM (1 unpaired), 2.83 (2), 3.87 (3), 4.90 (4), 5.92 (5).

Question 3. Which one of the following is wrongly matched ? (Karnataka CET 2010)
(a) [Cu(NH₃)₄]²⁺ – square planar
(b) [Ni(CO)₄] – neutral ligand
(c) [Fe(CN)₆]³⁻ – sp³d²
(d) [Co(en)₃]³⁺ – follows EAN rule
Answer:
Ans. (c)
[Fe(CN)₆]³⁻ is an inner orbital octahedral complex with the hybridisation d²sp³.

Teacher's Note:
1. CN⁻ is a strong-field ligand, so cyano complexes of Fe are inner orbital (d²sp³), low spin complexes.

Question 4. Complete removal of both the axial ligands (along the z-axis) from an octahedral complex leads to which of the following splitting patterns ? (relative orbital energies not on scale; orbitals listed from highest to lowest energy) (JEE Main 2019)
(a) d_x²−y² > d_xy > d_z² > d_xz, d_yz
(b) d_z² > d_x²−y² > d_xz, d_yz > d_xy
(c) d_x²−y² > d_z² > d_xy > d_xz, d_yz
(d) d_x²−y² > d_z² > d_xz, d_yz > d_xy
Answer:
Ans. (a)
Removing both axial ligands gives a square planar complex. Orbitals with a z component (d_z², d_xz, d_yz) are stabilised, while orbitals in the xy plane face the four remaining ligands.
d_x²−y² points directly at the ligands and is highest; d_xy lies in the same plane (between ligands) and comes next; then d_z², and the degenerate pair d_xz, d_yz is lowest.

Teacher's Note:
1. In square planar splitting, d_xy lies above d_z² because it lies in the ligand plane; d_xz/d_yz stay degenerate and lowest.

Question 5. Which of the following complex has minimum magnitude of Δ₀ ? (DPMT 2010, Karnataka CET 2015)
(a) [Cr(CN)₆]³⁻
(b) [Co(NH₃)₆]³⁺
(c) [CoCl₆]³⁻
(d) [Cr(H₂O)₆]³⁺
Answer:
Ans. (c)
Δ₀ depends on the strength of the ligand. In the spectrochemical series Cl⁻ < H₂O < NH₃ < CN⁻.
Cl⁻ is the weakest ligand, so [CoCl₆]³⁻ has the minimum Δ₀.

Teacher's Note:
1. Learn the short spectrochemical order I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < CN⁻ < CO.

Question 6. Transition metal complex with highest value of crystal field splitting (Δ₀) will be (JEE Main 2022)
(a) [Cr(H₂O)₆]³⁺
(b) [Mo(H₂O)₆]³⁺
(c) [Fe(H₂O)₆]³⁺
(d) [Os(H₂O)₆]³⁺
Answer:
Ans. (d)
All complexes have the same ligand (H₂O) and the same charge (+3) on the metal ion.
Δ₀ increases on going down a group (3d < 4d < 5d) because the larger 5d orbitals interact more strongly with the ligands. Os is a 5d metal, so [Os(H₂O)₆]³⁺ has the highest Δ₀.

Teacher's Note:
1. For the same ligand and charge, Δ₀ rises by roughly 30–50% from 3d to 4d and again to 5d.

Question 7. What is the correct electronic configuration of the central atom in K₄[Fe(CN)₆] based on crystal field theory ? (NEET 2019)
(a) e⁴ t₂²
(b) t_2g⁴ e_g²
(c) t_2g⁶ e_g⁰
(d) e³ t₂³
Answer:
Ans. (c)
In K₄[Fe(CN)₆], Fe is +2, i.e., 3d⁶. CN⁻ is a strong field ligand (Δ₀ > P), so all six electrons pair up in the lower t_2g set: t_2g⁶ e_g⁰.

Teacher's Note:
1. e and t₂ labels belong to tetrahedral splitting; an octahedral complex uses t_2g and e_g.

Question 8. The electronic spectrum of [Ti(H₂O)₆]³⁺ shows a single broad peak with a maximum at 20300 cm⁻¹. The crystal field stabilization energy (CFSE) of the complex ion, in kJ mol⁻¹, is (1 kJ mol⁻¹ = 83.7 cm⁻¹) (JEE Main 2020)
(a) 145.5
(b) 97
(c) 242.5
(d) 83.7
Answer:
Ans. (b)
The absorption maximum gives Δ₀ = 20300 cm⁻¹.
Ti³⁺ = 3d¹ = t_2g¹, so CFSE = −0.4 Δ₀ = −0.4 × 20300 = −8120 cm⁻¹
\( = \frac{8120}{83.7} \) kJ mol⁻¹ = 97 kJ mol⁻¹ (magnitude)

Teacher's Note:
1. For a d¹ ion the single d–d band directly gives Δ₀; CFSE is only 0.4 of it.

Question 9. Low spin complex of d⁶-cation in an octahedral field will have the following energy : (AIPMT Main 2012)
(a) \( -\frac{2}{5}\Delta_0 + 2P \)
(b) \( -\frac{2}{5}\Delta_0 + P \)
(c) \( -\frac{12}{5}\Delta_0 + P \)
(d) \( -\frac{12}{5}\Delta_0 + 3P \)
Answer:
Ans. (d)
Low spin d⁶ : t_2g⁶ e_g⁰
Energy \( = 6\times\left(-\frac{2}{5}\Delta_0\right) + 3P = -\frac{12}{5}\Delta_0 + 3P \) (three electron pairs in t_2g)

Teacher's Note:
1. Here the energy counts all three pairs in t_2g⁶; some books count only the extra pairs compared with the free ion, so read the options carefully.

Question 10. Jahn-Teller effect is not observed in high spin complexes of (NEET Phase II 2016)
(a) d⁷
(b) d⁸
(c) d⁴
(d) d⁹
Answer:
Ans. (b)
Jahn-Teller distortion occurs when a degenerate set of orbitals is unevenly filled.
High spin d⁸ = t_2g⁶ e_g²: both sets are symmetrically filled, so no Jahn-Teller effect. d⁴ (e_g¹), d⁷ (t_2g⁵) and d⁹ (e_g³) are unevenly filled.

Teacher's Note:
1. Strong Jahn-Teller distortion is seen for an uneven e_g set: high spin d⁴ and d⁹ (e.g. Cu²⁺).

Question 11. Two complexes [Cr(H₂O)₆Cl₃] (A) and [Cr(NH₃)₆] Cl₃ (B) are violet and yellow coloured respectively. The incorrect statement regarding them is (JEE Main 2019)
(a) Δ_o value of (A) is less than that of (B)
(b) Δ_o values of (A) and (B) are calculated from the energies of violet and yellow light
(c) Both absorb energies corresponding to their complementary colours
(d) Both are paramagnetic with three unpaired electrons
Answer:
Ans. (b)
The colour seen is the complementary colour of the light absorbed. Δ_o is calculated from the energy of the light absorbed, not from the energy of the colour transmitted (violet and yellow). Hence (b) is incorrect.
H₂O is a weaker ligand than NH₃, so Δ_o(A) < Δ_o(B); both contain Cr³⁺ (3d³) with three unpaired electrons.

Teacher's Note:
1. A violet complex absorbs yellow-green light and a yellow complex absorbs violet-blue light; always use the absorbed wavelength in E = hc/λ.

Question 12. The metal d-orbitals that are directly facing the ligands in K₃[Co(CN)₆] are : (JEE Main 2019)
(a) d_xz, d_yz and d_z²
(b) d_xy, d_xz and d_yz
(c) d_xy and d_x²−y²
(d) d_x²−y² and d_z²
Answer:
Ans. (d)
In an octahedral complex the six ligands approach along the x, y and z axes. The e_g orbitals, d_x²−y² and d_z², have their lobes along the axes and so directly face the ligands.

Teacher's Note:
1. t_2g orbitals (d_xy, d_yz, d_xz) point between the axes, which is why they are lower in energy in an octahedral field.

Question 13. The magnitude of crystal field stabilization energy (CFSE or Δ_t) in tetrahedral complexes is considerably less than in the octahedral field. Because (DPMT 2009)
(a) There are only four ligands instead of six so the ligand field is only 2/3 the size hence the Δ_t is only 2/3 the size
(b) The direction of the orbitals does not coincide with the direction of the ligands. This reduces the crystal field stabilization energy (Δ_t) by further 2/3.
(c) Both points (a) and (b) are correct.
(d) Both points (a) and (b) are wrong.
Answer:
Ans. (c)
Both factors act together: fewer ligands (4 instead of 6) reduce the field to 2/3, and the ligands do not approach along the orbital directions, reducing it by a further 2/3.
Hence \( \Delta_t = \frac{2}{3}\times\frac{2}{3}\Delta_0 = \frac{4}{9}\Delta_0 \).

Teacher's Note:
1. Because Δ_t ≈ 4/9 Δ₀ is usually smaller than the pairing energy, tetrahedral complexes are almost always high spin.

Question 14. The complex ion that will lose its crystal field stabilization energy upon oxidation of its metal to + 3 state is
(Phen = 1,10-phenanthroline, a bidentate ligand with two ring N donor atoms; ignore pairing energy) (JEE Main 2019)
(a) [Co (phen)₃]²⁺
(b) [Zn (phen)₃]²⁺
(c) [Fe (phen)₃]²⁺
(d) [Ni (phen)₃]²⁺
Answer:
Ans. (c)
Phen is a strong field ligand. For Fe²⁺ (3d⁶, t_2g⁶) CFSE = −2.4 Δ₀; on oxidation to Fe³⁺ (3d⁵, t_2g⁵) CFSE = −2.0 Δ₀. So CFSE is lost.
Co²⁺ (t_2g⁶e_g¹, −1.8 Δ₀) → Co³⁺ (t_2g⁶, −2.4 Δ₀) and Ni²⁺ (−1.2 Δ₀) → Ni³⁺ (−1.8 Δ₀) gain CFSE; Zn²⁺ (d¹⁰) has zero CFSE.

Teacher's Note:
1. t_2g⁶ (low spin d⁶) has the maximum possible CFSE, so any change away from it loses stabilisation.

Question 15. Correct increasing order for the wavelength of absorption in the visible region for the complexes of Co³⁺ is (NEET 2017)
(a) [Co(en)₃]³⁺, [Co(NH₃)₆]³⁺, [Co(H₂O)₆]³⁺
(b) [Co(H₂O)₆]³⁺, [Co(en)₃]³⁺, [Co(NH₃)₆]³⁺
(c) [Co(H₂O)₆]³⁺, [Co(NH₃)₆]³⁺, [Co(en)₃]³⁺
(d) [Co(NH₃)₆]³⁺, [Co(en)₃]³⁺, [Co(H₂O)₆]³⁺
Answer:
Ans. (a)
Ligand strength: H₂O < NH₃ < en. Stronger ligand → larger Δ₀ → higher energy absorbed → shorter wavelength.
Hence increasing order of wavelength: [Co(en)₃]³⁺ < [Co(NH₃)₆]³⁺ < [Co(H₂O)₆]³⁺.

Teacher's Note:
1. Wavelength absorbed runs opposite to ligand strength, because Δ₀ = hc/λ.

Question 16. Three complexes [CoCl(NH₃)₅]²⁺ (I), [Co(NH₃)₅ H₂O]³⁺ (II) and [Co(NH₃)₆]³⁺ (III) absorb light in the visible region. The correct order of the wavelengths of the light absorbed by them is (JEE Main 2021)
(a) III > I > II
(b) I > II > III
(c) II > I > III
(d) III > II > I
Answer:
Ans. (b)
The complexes differ in one ligand: Cl⁻ (I), H₂O (II) and NH₃ (III). Ligand strength: Cl⁻ < H₂O < NH₃, so Δ₀: I < II < III.
Wavelength absorbed is inversely related to Δ₀, so the order of wavelengths is I > II > III.

Teacher's Note:
1. When only one ligand differs, compare just that ligand's position in the spectrochemical series.

Question 17. The order of energy absorbed which is responsible for the colour of complexes
(A) [Ni(H₂O)₂(en)₂]²⁺ ; (B) [Ni(H₂O)₄(en)]²⁺ and (C) [Ni(en)₃]²⁺ is (NEET 2022)
(a) (A) > (B) > (C)
(b) (C) > (B) > (A)
(c) (C) > (A) > (B)
(d) (B) > (A) > (C)
Answer:
Ans. (c)
en is a stronger field ligand than H₂O. As H₂O molecules are replaced by en, Δ₀ and hence the energy absorbed increases.
Number of en: C (3) > A (2) > B (1). Hence energy absorbed: (C) > (A) > (B).

Teacher's Note:
1. More strong-field ligands mean larger splitting, so energy absorbed rises and the absorbed wavelength falls.

Question 18. Which of the following compounds is not yellow coloured ? (JEE Main 2015)
(a) (NH₄)₃ [As (Mo₃O₁₀)₄]
(b) BaCrO₄
(c) Zn₂ [Fe(CN)₆]
(d) K₃ [Co(NO₂)₆]
Answer:
Ans. (c)
Ammonium arsenomolybdate, barium chromate and potassium cobaltinitrite are all yellow precipitates.
Zn₂[Fe(CN)₆] contains Fe²⁺ (low spin t_2g⁶) and Zn²⁺ (d¹⁰); it is white (bluish-white), not yellow.

Teacher's Note:
1. These precipitates come from salt analysis: yellow BaCrO₄ (Ba²⁺), yellow K₃[Co(NO₂)₆] (K⁺/Co²⁺) and white Zn₂[Fe(CN)₆] (Zn²⁺).

Question 19. Which one of the following ligand is capable of forming a low spin as well as a high spin complex ?
(a) CO
(b) F⁻
(c) NH₃
(d) CN⁻
Answer:
Ans. (c)
CO and CN⁻ are very strong ligands and always give low spin complexes; F⁻ is a weak ligand and gives high spin complexes.
NH₃ lies in the middle of the spectrochemical series: it gives low spin [Co(NH₃)₆]³⁺ but high spin [Co(NH₃)₆]²⁺, depending on the metal ion.

Teacher's Note:
1. Whether pairing occurs depends on both Δ₀ and P; borderline ligands like NH₃ can go either way with different metal ions.

Question 20. Which of the following shall form an octahedral complex ?
(a) d⁴ (low spin)
(b) d⁸ (high spin)
(c) d⁶ (low spin)
(d) all of these.
Answer:
Ans. (d)
d⁴ (low spin): t_2g⁴, two inner d orbitals free → d²sp³, octahedral, e.g. [Mn(CN)₆]³⁻.
d⁸ (high spin): outer orbital sp³d², octahedral, e.g. [Ni(NH₃)₆]²⁺.
d⁶ (low spin): t_2g⁶ → d²sp³, octahedral, e.g. [Co(NH₃)₆]³⁺.
Hence all of these form octahedral complexes.

Teacher's Note:
1. Test each option with a known example before choosing a single answer; here every configuration has a familiar octahedral complex.

One or More Correct Answers

Question 21. Identify the complexes which are expected to be coloured.
(a) [Ti (NO₃)₄]
(b) [Cu (NCCH₃)₄]⁺ BF₄⁻
(c) [Cr (NH₃)₆]³⁺ 3 Cl⁻
(d) K₃ [VF₆].
Answer:
Ans. (c, d)
Only those complexes are coloured in which the central metal ion has incompletely filled d-subshell.
Ti(NO₃)₄ : Ti⁴⁺, 3d⁰ 4s⁰ → colourless
[Cu (NCCH₃)₄]⁺ BF₄⁻ : Cu⁺, 3d¹⁰ 4s⁰ → colourless
[Cr (NH₃)₆]³⁺ 3 Cl⁻ : Cr³⁺, 3d³ 4s⁰ → coloured
K₃ [VF₆] : V³⁺, 3d² 4s⁰ → coloured

Teacher's Note:
1. Find the oxidation state first; then d⁰ or d¹⁰ means colourless.

Question 22. The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are) (JEE Advanced 2013)
(a) [Cr(NH₃)₅Cl] Cl₂ and [Cr(NH₃)₄Cl₂] Cl
(b) [Co(NH₃)₄Cl₂]⁺ and [Pt(NH₃)₂(H₂O) Cl]⁺
(c) [CoBr₂Cl₂]²⁻ and [PtBr₂Cl₂]²⁻
(d) [Pt(NH₃)₃(NO₃)] Cl and [Pt(NH₃)₃Cl] Br
Answer:
Ans. (b, d)
Each complex ion in the pair [Co(NH₃)₄Cl₂]⁺, [Pt(NH₃)₂(H₂O)Cl]⁺ shows geometrical isomerism. Each complex in the pair [Pt(NH₃)₃(NO₃)] Cl, [Pt(NH₃)₃Cl] Br shows ionization isomerism. In other pairs, the two complexes/ions do not show the same type of isomerism.

Teacher's Note:
1. Tetrahedral [CoBr₂Cl₂]²⁻ shows no geometrical isomerism, while square planar [PtBr₂Cl₂]²⁻ does.

Question 23. The correct option(s) regarding the complex [Co(en)(NH₃)₃(H₂O)]³⁺, en = H₂N CH₂CH₂NH₂, is (are) (JEE Advanced 2018)
(a) it has two geometrical isomers
(b) it will have three geometrical isomers if bidentate ‘en’ is replaced by two cyanide ligands
(c) it is paramagnetic
(d) it absorbs light at longer wavelength as compared to [Co(en)(NH₃)₄]³⁺
Answer:
Ans. (a, b, d)
(a) [Co(en)(NH₃)₃(H₂O)]³⁺ has 2 geometrical isomers: fac-type (the three NH₃ on one face, H₂O trans to an en N) and mer-type (the three NH₃ on a meridian, H₂O trans to NH₃).
(b) If ‘en’ is replaced by 2 CN⁻, we will get [Co(CN)₂(NH₃)₃(H₂O)]⁺. It will have 3 geometrical isomers (fac-NH₃; mer-NH₃ with CN⁻ trans to each other; mer-NH₃ with CN⁻ cis to each other).
(c) As en is strong ligand, in [Co(en)(NH₃)₃(H₂O)]³⁺, pairing will occur. Hence it will be diamagnetic and not paramagnetic.
(d) As H₂O is weaker ligand than NH₃, it will absorb longer wavelength as compared to [Co(en)(NH₃)₄]³⁺.

Teacher's Note:
1. Co³⁺ with N-donor ligands is low spin t_2g⁶, so it is diamagnetic.

Passage-Based Questions

Question 24. Comprehension 2: A metal complex having the composition Cr (NH₃)₄ Cl₂ Br has been isolated in two forms A and B. The form A reacts with AgNO₃ to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia.
The spin only value of the magnetic moment of the complex is
(a) \( \sqrt{3} \) B.M.
(b) \( \sqrt{8} \) B.M.
(c) \( \sqrt{15} \) B.M.
(d) Zero.
Answer:
Ans. (c)
As there are three unpaired electrons,
spin magnetic moment (μ) \( = \sqrt{n(n+2)} \) B.M. \( = \sqrt{3\times5} \) B.M. \( = \sqrt{15} \) B.M.

Teacher's Note:
1. √15 ≈ 3.87 BM is the value to remember for any d³ ion such as Cr³⁺.

Question 25. Comprehension 3: The coordination number of Ni²⁺ is 4
NiCl₂ + KCN (excess) → A (cyano complex)
A + conc. HCl (excess) → B (chloro complex) (IIT 2006)
The IUPAC names of A and B are
(a) Potassium tetracyanonickelate(II), potassium tetrachloronickelate(II)
(b) Tetracyanopotassium nickelate(II), tetrachloro-potassiumnickelate(II)
(c) Tetracyanonickel(II), tetrachloronickel(II)
(d) Potassium tetracyanonickel(II), potassium tetrachloronickel(II)
Answer:
Ans. (a)
NiCl₂ + 4 KCN → K₂[Ni(CN)₄] (A, Potassium tetracyanonickelate(II)) + 2 KCl
K₂[Ni(CN)₄] + 4 HCl (excess) → K₂[NiCl₄] (B, Potassium tetrachloronickelate(II)) + 4 HCN

Teacher's Note:
1. Name the cation first, then the anionic complex with the metal ending in “-ate”.

Matrix Match Questions

Teacher's Note:
1. Classify each complex by type ([M(AA)₂B₂], [M(AA)₃], [MA₃B₃], tetrahedral MA₂B₂) and recall the standard result for that type.

Teacher's Note:
1. Every complex here matches exactly one shape and one magnetic property; check both for each row.

Integer and Numerical Answer Questions

Question 28. The number of geometrical isomers possible for the complex [CoL₂Cl₂]⁻ (L = H₂NCH₂CH₂O⁻) is (JEE Advanced 2016)
Answer:
Ans. 5
L is an unsymmetrical bidentate ligand (N and O donors), so the complex is of the type [M(AB)₂C₂].
With the two Cl trans to each other, the two L ligands lie in one plane: N trans to N (I) or N trans to O (II).
With the two Cl cis to each other: the two N atoms trans to each other (III), the two O atoms trans to each other (IV), or N trans to O (V).
Hence, total number of geometrical isomers = 5.

Teacher's Note:
1. For [M(AB)₂C₂], first fix C–C trans or cis, then list the possible trans partners of each donor atom.

Question 29. Among the species given below, the total number of diamagnetic species is...........
H atom, NO₂ monomer, O₂⁻ (superoxide), dimeric sulphur in vapour phase, Mn₃O₄, (NH₄)₂[FeCl₄], (NH₄)₂[NiCl₄], K₂MnO₄, K₂CrO₄. (JEE Advanced 2018)
Answer:
Ans. 1
Paramagnetic species are H atom, NO₂ monomer, O₂⁻ → odd electron species.
Dimeric sulphur in vapour phase → According to Molecular Orbital Theory (two unpaired electrons, like O₂)
Mn₃O₄, (NH₄)₂[FeCl₄], (NH₄)₂[NiCl₄], K₂MnO₄ → Unpaired electron in d-orbital of central atom.
Diamagnetic
K₂CrO₄ → No unpaired electron in d-orbital of central atom (Cr⁶⁺, 3d⁰).

Teacher's Note:
1. S₂ is isoelectronic in valence shell with O₂, so it is paramagnetic; Mn⁶⁺ in manganate is d¹.

Question 30. Total number of cis N–Mn–Cl bond angles (that is Mn–N and Mn–Cl bonds in cis-positions) present in a molecule of cis-[Mn(en)₂Cl₂] complex is .......... (en = NH₂CH₂CH₂NH₂) (JEE Advanced 2019)
Answer:
Ans. 6
In cis-[Mn(en)₂Cl₂], label the two Cl atoms Cl_(a), Cl_(b) and the four N atoms N₍₁₎–N₍₄₎. Each Cl is trans to one N and cis to the other three.
cis angles: N₍₁₎–Mn–Cl_(a), N₍₁₎–Mn–Cl_(b), N₍₂₎–Mn–Cl_(a), N₍₃₎–Mn–Cl_(b), N₍₄₎–Mn–Cl_(a), N₍₄₎–Mn–Cl_(b)
Total number of cis N–Mn–Cl bond angles = 6

Teacher's Note:
1. Quick count: 4 N × 2 Cl = 8 N–Mn–Cl angles, minus the 2 that are trans (180°) = 6.

Question 31. Number of bridging CO ligands in [Mn₂(CO)₁₀] is ......... . (JEE Main 2021)
Answer:
Ans. 0
In Mn₂(CO)₁₀, the two Mn(CO)₅ units are held together only by a Mn–Mn bond; all ten CO ligands are terminal. Hence, number of bridging CO = 0.

Teacher's Note:
1. Compare Fe₂(CO)₉ (three bridging CO) and Co₂(CO)₈ (two bridging CO in the solid); Mn₂(CO)₁₀ has none.

Question 32. The equivalents of ethylene diamine required to replace the neutral ligands from the coordination sphere of the trans-complex of Co.4 NH₃ is .......... . (JEE Main 2021)
Answer:
Ans. 2
The trans-complex is trans-[Co(NH₃)₄Cl₂] Cl. Its four neutral NH₃ ligands can be replaced by two ‘en’ ligands giving [Co(en)₂Cl₂] Cl.

Teacher's Note:
1. Each en is didentate and replaces two monodentate NH₃ ligands.

Assertion–Reason Questions

Question 33. Assertion: EDTA is a hexadentate ligand.
Reason: Denticity of a ligand is given by number of lone pairs donated to central atom by a ligand. (AIIMS 2012)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
R is the correct explanation of A: EDTA donates six lone pairs (2 N + 4 O) to the central atom.

Teacher's Note:
1. Denticity counts donor atoms actually bound to the metal, not the total lone pairs present on the ligand.

Question 34. Assertion: [Co(NO₂)₃(NH₃)₃] does not show optical isomerism.
Reason: It has a plane of symmetry. (AIIMS 2006)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
R is the correct explanation of A: both the fac and mer isomers of [MA₃B₃] have a plane of symmetry.

Teacher's Note:
1. A molecule with a mirror plane is superimposable on its mirror image, so it cannot be optically active.

Question 35. Assertion: Both [Ni(CN)₄]²⁻ and [NiCl₄]²⁻ have same shape and same magnetic behaviour.
Reason: Both are square planar and diamagnetic.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (d)
Correct A. [Ni(CN)₄]²⁻ is square planar and diamagnetic whereas [NiCl₄]²⁻ is tetrahedral and paramagnetic.
Correct R. [Ni(CN)₄]²⁻ is square planar while [NiCl₄]²⁻ is tetrahedral.

Teacher's Note:
1. Same metal ion and same coordination number can still give different shapes when the ligand field strength differs.

Question 36. Assertion: NF₃ is weaker ligand than N(CH₃)₃.
Reason: NF₃ ionizes to give F⁻ ions in aqueous solution. (AIIMS 2003)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (c)
Correct R. F is highly electronegative. It attracts the electrons of N towards itself and hence are not easily available on N, making it a weaker donor or ligand.

Teacher's Note:
1. Electron-withdrawing groups (F) reduce the donor power of N, while electron-releasing CH₃ groups increase it.